Answer:
i) 3.514 s, ii) 5.692 m/s
Explanation:
i) We can use Newton's second law of motion to find out how long does it take for the Eagle to touch down.
as the equation says for free-falling
h = ut +0.5gt^2
Here, h = 10 m, g = acceleration due to gravity = 1.62 m/s^2( on moon surface)
initial velocity u = 0
10 = 0.5×1.62t^2
t = 3.514 seconds
Therefore, it takes t = 3.514 seconds for the Eagle to touch down.
ii) use Newton's 1st equation of motion to calculate the velocity of the lunar module when it hits the surface of the moon
v = u + gt
v = 0+ 1.62×3.514
v= 5.692 m/s
Answer:
5 L
Explanation:
Ideal gas law:
PV = nRT
If P, n, and R are constant, then:
n₁R/P₁ = n₂R/P₂
Using ideal gas law, we can rewrite this as:
V₁/T₁ = V₂/T₂
This is known as Charles' law.
Plugging in values:
10 L / 546 K = V / 273 K
V = 5 L
The answer is a Mid-Ocean ridge.
Answer:
option (d) 7.1 kN
Explanation:
Given:
Mass of the car, m = 1600 kg
Acceleration of the car, a = 1.5 m/s²
Coefficient of kinetic friction = 0.3
let the tension be 'T'
Now,
ma = T - f .................(1)
where f is the frictional force
also,
f = 0.3 × mg
where g is the acceleration due to the gravity
thus,
f = 0.3 × 1600 × 9.81 =
therefore,
equation 1 becomes
1600 × 1.5 = T - 4708.8
or
T = 2400 + 4708.8
or
T = 7108.8 N
or
T = 7.108 kN
Hence,
The correct answer is option (d) 7.1 kN
=0.6
